Trigonometry/For Enthusiasts/Doing without Sine

The Idea

We know that:

$\sin^2\theta = 1 - \cos^2\theta\,$

So do we really need the $\sin\,$ function?

Or put another way, could we have worked out all our interesting formulas for things like $\cos( a+b )\,$ in terms just of $\cos\,$ and then derived every formula that has a $\sin\,$ in it from that?

The answer is yes.

We don't need to have one geometric argument for $\cos(a+b)\,$ and then do another geometric argument for $\sin(a+b)\,$. We could get our formulas for $\sin\,$ directly from formulas for $\cos\,$

Angle Addition and Subtraction formulas

To find a formula for $\cos(\theta_1+\theta_2)\,$ in terms of $\cos\left( \theta_1\right)$ and $\cos\left( \theta_2\right)$: construct two different right angle triangles each drawn with side $c\,$ having the same length of one, but with $\theta_1 \ne \theta_2$, and therefore angle $\psi_1 \ne \psi_2$. Scale up triangle two so that side $a_2\,$ is the same length as side $c_1\,$. Place the triangles so that side $c_1\,$ is coincidental with side $a_2\,$, and the angles $\theta_1\,$ and $\theta_2\,$ are juxtaposed to form angle $\theta_3=\theta_1+\theta_2\,$ at the origin. The circumference of the circle within which triangle two is embedded (circle 2) crosses side $a_1\,$ at point $g\,$, allowing a third right angle to be drawn from angle $\varphi_2$ to point $g\,$ . Now reset the scale of the entire figure so that side $c_2\,$ is considered to be of length 1. Side $a_2\,$ coincidental with side $c_1\,$ will then be of length $\cos\left( \theta_1\right)$, and so side $a_1\,$ will be of length $\cos\left( \theta_1\right) \cdot \cos\left( \theta_2\right)$ in which length lies point $g\,$. Draw a line parallel to line $a_1\,$ through the right angle of triangle two to produce a fourth right angle triangle, this one embedded in triangle two. Triangle 4 is a scaled copy of triangle 1, because:

 (1) it is right angled, and 
 (2) $\theta_4 + (\pi - \frac{\pi}{2} - \theta_1 - \theta_2) = \varphi_2 = \pi - \frac{\pi}{2} - \theta_2 \Rightarrow \theta_4 = \theta_1$. 

The length of side $b_4\,$ is $cos(\varphi_2)cos(\varphi_4) = cos(\varphi_2)cos(\varphi_1)$ as $\theta_4\, = \theta_1\,$. Thus point $g\,$ is located at length:

$cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - cos(\varphi_1)cos(\varphi_2),$ where $\theta_1+\varphi_1 = \theta_2+\varphi_2 = \frac{\pi}{2}$

giving us the "Cosine Angle Sum Formula".

Proof that angle sum formula and double angle formula are consistent

We can apply this formula immediately to sum two equal angles:

   $cos(2\theta) = cos(\theta+\theta) = cos(\theta)cos(\theta) - cos(\varphi)cos(\varphi)  = cos(\theta)^{2} - cos(\varphi)^{2}$           (I)
    where $\theta+\varphi = \frac{\pi}{2}$

From the theorem of Pythagoras we know that:

  $a^2 + b^2 = c^2\,$

in this case:

   $cos(\theta)^2 + cos(\varphi)^2  = 1^2$$\Rightarrow cos(\varphi)^2 = 1 - cos(\theta)^2$
   where $\theta+ \varphi = \frac{\pi}{2}$

Substituting into (I) gives:

   $cos(2\theta)=cos(\theta)^2-cos(\varphi)^2$$= cos(\theta)^2 - (1 - cos(\theta)^2)\,$$= 2 cos(\theta)^2 - 1\,$
   where $\theta+\varphi = \frac{\pi}{4}$

which is identical to the "Cosine Double Angle Sum Formula":

   $2 cos(\delta)^2 - 1 = cos(2\delta)\,$

Pythagorean identity

Armed with this definition of the $sin()\,$ function, we can restate the Theorem of Pythagoras for a right angled triangle with side c of length one, from:

    $cos(\theta)^2 + cos(\varphi)^2 = 1^2$  where $\theta+\varphi = \frac{\pi}{2}$

to:

    $cos(\theta)^2 + sin(\theta)^2 = 1\,$

We can also restate the "Cosine Angle Sum Formula" from:

 $cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - cos(\varphi_1)cos(\varphi_2)$ where $\theta_1+\varphi_1 = \theta_2+\varphi_2 = \frac{\pi}{2}$

to:

 $cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2)\,$

Sine Formulas

The price we have to pay for the notational convenience of this new function $sin()\,$ is that we now have to answer questions like: Is there a "Sine Angle Sum Formula". Such questions can always be answered by taking the $cos()\,$ form and selectively replacing $cos(\theta)^2\,$ by $1 - sin(\theta)^2\,$ and then using algebra to simplify the resulting equation. Applying this technique to the "Cosine Angle Sum Formula" produces:

   $cos(\theta_1+\theta_2) = cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2)\,$$\Rightarrow cos(\theta_1+\theta_2)^2 = (cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2))^2\,$$\Rightarrow 1 - cos(\theta_1+\theta_2)^2 = 1 - (cos(\theta_1)cos(\theta_2) - sin(\theta_1)sin(\theta_2))^2\,$$\Rightarrow sin(\theta_1+\theta_2)^2     = 1 - (cos(\theta_1)^2cos(\theta_2)^2 + sin(\theta_1)^2sin(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,$
   -- Pythagoras on left, multiply out right hand side

                     $= 1 - (cos(\theta_1)^2(1-sin(\theta_2)^2) + sin(\theta_1)^2(1-cos(\theta_2)^2) - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,$
                     -- Carefully selected Pythagoras again on the left hand side

                     $= 1 - (cos(\theta_1)^2-cos(\theta_1)^2sin(\theta_2)^2 + sin(\theta_1)^2-sin(\theta_1)^2cos(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,$
                     -- Multiplied out

                     $= 1 - (1 -cos(\theta_1)^2sin(\theta_2)^2 -sin(\theta_1)^2cos(\theta_2)^2 - 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2))\,$
                     -- Carefully selected Pythagoras 

                     $= cos(\theta_1)^2sin(\theta_2)^2+sin(\theta_1)^2cos(\theta_2)^2 + 2cos(\theta_1)cos(\theta_2)sin(\theta_1)sin(\theta_2)\,$
                     -- Algebraic simplification

                     $=(cos(\theta_1)sin(\theta_2)+sin(\theta_1)cos(\theta_2))^2\,$

taking the square root of both sides produces the "Sine Angle Sum Formula"

$\Rightarrow sin(\theta_1+\theta_2) = cos(\theta_1)sin(\theta_2) + sin(\theta_1)cos(\theta_2)$

We can use a similar technique to find the "Sine Half Angle Formula" from the "Cosine Half Angle Formula":

$cos \left (\frac{\theta}{2} \right ) = \sqrt{\frac{1 + cos(\theta)}{2}}$

We know that $1 - cos \left ( \frac{\theta}{2} \right ) ^2 = sin \left ( \frac{\theta}{2} \right ) ^2$, so squaring both sides of the "Cosine Half Angle Formula" and subtracting from one:

 $\Rightarrow 1 - cos \left ( \frac{\theta}{2} \right ) ^2 =       1 - \frac{1+cos(\theta)}{2}$$\Rightarrow     sin \left ( \frac{\theta}{2} \right ) ^2  =            \frac{1-cos(\theta)}{2}$$\Rightarrow     sin \left ( \frac{\theta}{2} \right ) = \sqrt{\frac{1-cos(\theta)}{2}}$

So far so good, but we still have a $\frac{cos(\theta)}{2}$ to get rid of. Use Pythagoras again to get the "Sine Half Angle Formula":

         $sin\left (\frac{\theta}{2} \right ) = \sqrt{\frac{1 - \sqrt{1 - sin(\theta)^2}}{2}}$

or perhaps a little more legibly as:

         $sin \left ( \frac{\theta}{2} \right  ) = \sqrt{\frac{1}{2}}\sqrt{1 - \sqrt{1 - sin(\theta)^2}}$